Problem 5 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The probability distribution of a random variable \(X\) is given. Compute the mean, variance, and standard deviation of \(X\). $$\begin{array}{lccccc}\hline \boldsymbol{x} & 430 & 480 & 520 & 565 & 580 \\ \hline \boldsymbol{P}(\boldsymbol{X}=\boldsymbol{x}) & .1 & .2 & .4 & .2 & .1 \end{array}$$

Step-by-Step Solution

Answer

The mean of the given probability distribution is 518, its variance is 1891, and its standard deviation is approximately 43.47.

Step 1: Calculate the mean

To find the mean, we can use the formula: \(E(X) = \sum x_i P(X=x_i)\). In our case, this will be: \(E(X) = 430(0.1) + 480(0.2) + 520(0.4) + 565(0.2) + 580(0.1)\) Evaluating this expression, we get: \(E(X) = 43 + 96 + 208 + 113 + 58 = 518\) So the mean of this probability distribution is 518.

Step 2: Calculate \(E(X^2)\)

We need to calculate \(E(X^2)\) for the variance formula. We can use the formula: \(E(X^2) = \sum x_i^2 P(X=x_i)\) In our case, this will be: \(E(X^2) = 430^2(0.1) + 480^2(0.2) + 520^2(0.4) + 565^2(0.2) + 580^2(0.1)\) Evaluating this expression, we get: \(E(X^2) = 18490 + 46080 + 108160 + 63845 + 33640 = 270215\) So, \(E(X^2) = 270215\).

Step 3: Calculate the variance

Now, we can use the formula for variance: \(Var(X) = E(X^2) - E(X)^2\) Plugging in the values we found in Steps 1 and 2, we get: \(Var(X) = 270215 - 518^2 = 270215 - 268324 = 1891\) So the variance of this probability distribution is 1891.

Step 4: Calculate the standard deviation

Finally, we can use the formula for standard deviation: \(SD(X) = \sqrt{Var(X)}\) Taking the square root of the variance we found in Step 3, we get: \(SD(X) = \sqrt{1891} \approx 43.47\) So the standard deviation of this probability distribution is approximately 43.47. In conclusion, the mean of the given probability distribution is 518, its variance is 1891, and its standard deviation is approximately 43.47.

Key Concepts

Mean

The mean, also known as the expected value, is a measure that gives us an average of a probability distribution. It tells us where most of the data is centered. To calculate the mean of a random variable, you multiply each possible value of the variable by its probability, then sum up all these products. For our random variable \(X\), the mean \(E(X)\) is calculated as follows:

• Multiply each value by its corresponding probability: \(430 \times 0.1\), \(480 \times 0.2\), \(520 \times 0.4\), \(565 \times 0.2\), \(580 \times 0.1\).
• Add all these products together: \(43 + 96 + 208 + 113 + 58\).
• The sum, which is 518, represents the mean of the distribution.

This means that the average value of \(X\) that we expect to get over many observations is 518.

Variance

Variance gives us an idea of how spread out the values of a random variable are around the mean. It measures the degree of variability and is crucial for understanding the distribution's consistency. A smaller variance implies the data points are clustered closely around the mean, whereas a larger variance indicates greater dispersal.To find the variance \(Var(X)\) of \(X\):

• First, calculate \(E(X^2)\) using the formula: \(E(X^2) = \sum x_i^2 P(X=x_i)\).
• For our values, this becomes \(430^2 \times 0.1 + 480^2 \times 0.2 + 520^2 \times 0.4 + 565^2 \times 0.2 + 580^2 \times 0.1\).
• The result of this calculation is 270215.
• The variance is then \(Var(X) = E(X^2) - (E(X))^2\), which gives us \(270215 - 518^2 = 1891\).

Therefore, the variance of this distribution is 1891, indicating how widely spread the distribution values are around the mean.

Standard Deviation

The standard deviation is perhaps one of the most closely watched indicators when it comes to understanding variability. It's the square root of the variance and provides insights into the average distance of each data point from the mean. Compared to variance, standard deviation is on the same scale as the data, making it intuitive to interpret.For our purposes, the standard deviation \(SD(X)\) is calculated as follows:

• Take the square root of the variance \(Var(X)\).
• In this scenario, \(SD(X) = \sqrt{1891} \approx 43.47\).

This tells us that on average, the values deviate from the mean by approximately 43.47 units. Understanding the standard deviation is essential for assessing the reliability and accuracy of the expected value (mean) in predicting outcomes.